Recurring decimals as fractions

Earlier I solved this question and it got me thinking.

Why do recurring decimals have a fractional equivalent?\text{Why do recurring decimals have a fractional equivalent?}

So I decided to work it out, this is what I got.

Let bb be a number with nn digits, it can have both leading and trailing zeros.

Every recurring decimal can now be represented by the decimal : 0.bbb0.bbb\ldots

To represent this as a fraction we need to see it as an infinite sum.

0.bbb=b10n+b102n+b103n+0.bbb\ldots = \frac{b}{10^{n}} + \frac{b}{10^{2n}} + \frac{b}{10^{3n}} + \ldots

Now that we have it as an infinite sum we can use the geometric series formula where

a=b10n,r=110na = \frac{b}{10^{n}}, r = \frac{1}{10^{n}}

x=0arx=a1r\sum\limits_{x = 0}^{\infty} ar^{x} = \frac{a}{1 - r}

Hence

x=0b10n(x+1)=b10n1110n\sum\limits_{x = 0}^{\infty}\frac{b}{10^{n(x + 1)}} = \frac{\frac{b}{10^{n}}}{1 - \frac{1}{10^{n}}}

b10n1110n=b10n10n110n=b10n10n10n1=b10n1\large\frac{\frac{b}{10^{n}}}{1 - \frac{1}{10^{n}}} = \frac{\frac{b}{10^{n}}}{\frac{10^{n} - 1}{10^{n}}} = \frac{b}{10^{n}} \cdot \frac{10^{n}}{10^{n} - 1} = \frac{b}{10^{n} - 1}

So that means that :

0.bbb=b10n10.bbb \ldots = \frac{b}{10^{n} - 1}

But what if there's some decimal places in front of the recurring part?

Let cc be a number with n1n_1 digits, it can also have leading and trailing zeros.

The decimal is now : 0.cbbb0.cbbb \ldots

Now we have to account for the decimal places between the decimal point and the recurring decimal. This is easy to do since we already have the equation for recurring decimals. The solution is simple - divide by 1010 to the power of the number of places (n1n_1).

0.cbbb= ?+b10n1(10n1)0.cbbb\ldots = \text{ ?} + \frac{b}{10^{n_1}(10^{n} - 1)}

The ?\text{?} represents the cc part of the decimal. It is simply the number cc divided by 1010 to the power of the number of decimal places (n1n_1) it takes up.

0.cbbb=c10n1+b10n1(10n1)0.cbbb\ldots = \frac{c}{10^{n_1}} + \frac{b}{10^{n_1}(10^{n} - 1)}

Let's simplify it a bit.

0.cbbb=b+c(10n1)10n1(10n1)0.cbbb\ldots = \frac{b + c(10^{n} - 1)}{10^{n_1}(10^{n} - 1)}

That's all for now, hope you found this note interesting.

Note by Jack Rawlin
3 years, 7 months ago

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Great! Could you add this to the Converting Repeating Decimals into Fractions wiki? Thanks!

Calvin Lin Staff - 3 years, 7 months ago

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